Question

Prove that the bisector of two adjacent angles of a parallelogram intersect at right angles.​

Answer 1

Question :-

• Prove that the bisector of two adjacent angles of a parallelogram intersect at right angles.

$\bf\underbrace\red{Answer:}$

Given :-

• A parallelogram ABCD in which bisectors of ∠DAB and ∠ABC meet at the point O.

To Prove :-

• ∠AOB = 90°.

Proof :-

Let ∠DAB = 2x and ∠ABC = 2y.

Since, ABCD is a parallelogram.

∴ ∠DAB + ∠ABC = 180° ( Sum of adjacent angles)

⇛ 2x + 2y = 180°

⇛ 2(x + y) = 180°

⇛ x + y = 90° ..........(1)

Now, In ∆ AOB,

Using angle sum property of triangle,

⇛ ∠OAB + ∠OBA + ∠AOB = 180°

⇛ x + y + ∠AOB = 180°

⇛ 90° + ∠AOB = 180° [Using (1) ]

⇛ ∠AOB = 180° - 90°

⇛ ∠AOB = 90°

$\bf\underbrace\red{Hence, \: Proved}$

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Additional Information

Definition of parallelogram :-

• If a quadrilateral has a pair of parallel opposite sides, then it’s a special polygon called Parallelogram.

The properties of a parallelogram are as follows:

• The opposite sides are parallel and congruent.

• The opposite angles are congruent.

• The consecutive angles are supplementary.

• If anyone of the angles is a right angle, then all the other angles will be the right angle.

• The two diagonals bisect each other.

• Each diagonal bisects the parallelogram into two congruent triangles.

• Sum of square of all the sides of parallelogram is equal to the sum of square of its diagonals. It is also called parallelogram law.